Which of the following will not show deflection from the path on passing through an electric field? Proton, ca

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Q. Which of the following will not show deflection from the path on passing through an electric field? Proton, cathode rays, electron, neutron.

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Concept understanding — De Broglie Wavelength

De Broglie Wavelength: When Particles Start Acting Like Waves

You already know that light behaves like a wave (interference, diffraction) and like a particle (photoelectric effect). That's wave-particle duality for light. De Broglie's radical idea in 1924 was: if light can be both, why can't matter be both too?

He proposed that every moving particle — an electron, a proton, even a cricket ball — has a wavelength associated with it. The faster it moves, the shorter that wavelength becomes.

The Intuition

Think of a wave on a string. Its wavelength is the distance between two consecutive crests. Now imagine an electron moving through space. De Broglie said that the electron's motion itself creates a "matter wave" — a wave of probability that guides where the electron is likely to be found.

Note

You never see this wavelength in everyday life because for large objects it's unimaginably tiny. A cricket ball moving at 30 m/s has a de Broglie wavelength of about $10^{-34}$ m — far smaller than an atomic nucleus. That's why macroscopic objects behave like particles.

The Precise Statement

The de Broglie wavelength $\lambda$ of a particle is given by:

\lambda = \frac{h}{p}

where:

$h$ is Planck's constant ( $6.626 \times 10^{-34} \ \text{J·s}$ )
$p$ is the momentum of the particle ( $p = mv$ for non-relativistic speeds)

Key point: The wavelength depends only on momentum, not on charge, mass, or any other property. A fast electron and a slow proton can have the same wavelength if their momenta are equal.

What This Means Physically

For electrons in atoms: The de Broglie wavelength of an electron in a hydrogen atom is roughly the size of the atom itself ( $\approx 10^{-10}$ m). This is why electrons form standing waves around the nucleus — only certain wavelengths "fit" into the orbit, which explains quantised energy levels.
For experiments: If you fire electrons through a crystal, they diffract just like X-rays. This was confirmed by Davisson and Germer in 1927 — a Nobel-winning experiment that proved de Broglie right.
For large objects: The wavelength is so small that wave behaviour is undetectable. A car moving at 100 km/h has $\lambda \approx 10^{-38}$ m — you'd need a slit smaller than an atom to see diffraction.

Watch out

A common mistake is to think the de Broglie wavelength is the size of the particle. It is not. It is the wavelength of the probability wave associated with the particle. The particle itself remains point-like.

Worked Example

Question: What is the de Broglie wavelength of an electron moving at $2.0 \times 10^6$ m/s? (Mass of electron $m_e = 9.11 \times 10^{-31}$ kg)

Solution:

First, find momentum:

p = mv = (9.11 \times 10^{-31})(2.0 \times 10^6) = 1.822 \times 10^{-24} \ \text{kg·m/s}

Then apply de Broglie's formula: …

Why this formula?

De Broglie Wavelength: Why Matter Has a Wave Nature

The idea that a moving particle has a wavelength is one of the most radical shifts in physics. It came from Louis de Broglie in 1924, who asked a simple question: if light — which we thought was a wave — can behave like a particle (the photon), then why can't a particle behave like a wave?

The Core Insight: Symmetry in Nature

De Broglie started from Einstein's relation for a photon. For light, the energy $E$ and momentum $p$ of a photon are linked to its wave properties — frequency $f$ and wavelength $\lambda$ — by:

E = h f \quad \text{and} \quad p = \frac{h}{\lambda}

where $h$ is Planck's constant. These are not arbitrary; they come from the fact that light is an electromagnetic wave, and Planck had already shown that energy comes in quanta $hf$ .

De Broglie's reasoning was a leap of symmetry: if nature treats light and matter on equal footing (as Einstein's special relativity suggests), then any moving particle should also have a wavelength associated with it. He proposed that the same relation holds for matter:

\lambda = \frac{h}{p}

where $p = mv$ is the momentum of the particle (for non-relativistic speeds). This is the de Broglie wavelength.

Why This Formula Makes Sense: A Simple Derivation

There is no rigorous "derivation" from first principles — de Broglie's hypothesis was a postulate. But we can see why it is plausible by combining two key ideas from relativity and quantum theory.

Step 1: Energy of a particle from relativity

For a particle with rest mass $m_0$ , the total energy in special relativity is:

E = \sqrt{p^2 c^2 + m_0^2 c^4}

For a photon, $m_0 = 0$ , so $E = pc$ . This matches the photon's wave relation $E = hf$ and $p = h/\lambda$ .

Step 2: Assume the same wave-particle duality for matter

If a massive particle also has a wave associated with it, then its energy should also be $E = hf$ , where $f$ is the frequency of the matter wave. Equating the relativistic energy with the quantum energy:

hf = \sqrt{p^2 c^2 + m_0^2 c^4}

For a particle moving at non-relativistic speeds ( $v \ll c$ ), the momentum $p = mv$ is small compared to $m_0 c$ , so we can expand:

E \approx m_0 c^2 + \frac{p^2}{2m_0}

The first term is rest energy, which is constant. The second term is kinetic energy $K = p^2/(2m)$ . The wave frequency $f$ then corresponds to the kinetic part (since rest energy doesn't contribute to motion). But the key relation we want is between wavelength and momentum.

Step 3: The wavelength from the wave speed

For any wave, the speed $v_{\text{wave}} = f \lambda$ . For a matter wave, de Broglie proposed that the wave speed equals the particle's speed $v$ (this is the phase velocity). So:

v = f \lambda

Now use $E = hf$ and $E = \frac{1}{2}mv^2$ (non-relativistic kinetic energy). Then:

f = \frac{E}{h} = \frac{mv^2}{2h}

Substitute into $v = f \lambda$ :

v = \frac{mv^2}{2h} \lambda \quad \Rightarrow \quad \lambda = \frac{2h}{mv}

This gives $\lambda = 2h/p$ , which is wrong by a factor of 2. The correct formula is $\lambda = h/p$ . …

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Particles with an electric charge experience a force in an electric field, causing them to deflect. Neutral particles, having no net charge, do not experience such a force and thus do not deflect. The neutron is electrically neutral.

The particle that will not show deflection is the neutron.

When a charged particle passes through an electric field, it experiences an electric force. This force causes the particle's path to change, leading to deflection. The direction and magnitude of this force depend on the charge of the particle and the strength and direction of the electric field.

The fundamental principle is that an electric field exerts a force only on charged objects. If a particle has no net electric charge, it will not experience an electric force when passing through an electric field, and therefore its path will remain unchanged.

\vec{F}

Let's analyze each option:

Proton:
- A proton is a subatomic particle found in the nucleus of an atom.
- It carries a positive elementary charge, $q = +e$ .
- Since it is charged, it will experience an electric force $\vec{F} = (+e)\vec{E}$ when passing through an electric field.
- This force will cause the proton to deflect from its original path. The deflection will be in the direction of the electric field.
Cathode rays:
- Cathode rays are streams of electrons. They are produced when a high voltage is applied across electrodes in a vacuum tube.
- Each electron in the cathode ray carries a negative elementary charge, $q = -e$ .
- Since cathode rays consist of charged particles (electrons), they will experience an electric force $\vec{F} = (-e)\vec{E}$ when passing through an electric field.
- This force will cause the cathode rays to deflect. The deflection will be in the direction opposite to the electric field.
Electron:
- An electron is a fundamental subatomic particle.
- It carries a negative elementary charge, $q = -e$ .
- Since it is charged, it will experience an electric force $\vec{F} = (-e)\vec{E}$ when passing through an electric field. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$M_1$
›Reveal solutionSolution
The photoelectric effect occurs only if the incident photon energy exceeds the metal’s work function. For λ = 310 nm, the photon energy is 4.0 eV; metals with work function > 4.0 eV (M₁, M₂, M₃) do not emit, so the correct option is (B).

Concept & Intuition

The photoelectric effect requires that the energy of an incoming photon be at least as large as the work function (the minimum energy needed to free an electron). If the photon energy is too low, no electrons are ejected, no matter how intense the light. Here we are given the wavelength of light; we convert it to energy (in eV) and compare with each metal’s work function. Metals whose work function is greater than the photon energy will show no photoelectric effect.

Step-by-step reasoning

Find the photon energy in eV The energy of a photon is $E = \frac{hc}{\lambda}$ . Use the convenient constant: $hc = 1240\ \text{eV·nm}$ (since $h = 4.1357\times10^{-15}\ \text{eV·s}$ and $c = 3\times10^8\ \text{m/s}$ ). For $\lambda = 310\ \text{nm}$ :

$E = \frac{1240\ \text{eV·nm}}{310\ \text{nm}} = 4.0\ \text{eV}.$

Compare with each work function

$M_1$ : work function = $4.8\ \text{eV}$ → $4.8 > 4.0$ → no emission

$M_2$ : work function = $4.3\ \text{eV}$ → $4.3 > 4.0$ → no emission

$M_3$ : work function = $4.75\ \text{eV}$ → $4.75 > 4.0$ → no emission

$M_4$ : work function = $3.75\ \text{eV}$ → $3.75 < 4.0$ → emission occurs …
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TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
$V_1$
›Reveal solutionSolution
The de Broglie wavelength of a charged particle accelerated through a potential $V$ depends on its mass and charge. Equating the wavelengths for a proton and an alpha particle gives the ratio $V_1 : V_2 = 8 : 1$ .

The de Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$ , where $p$ is its momentum. When a charged particle is accelerated from rest through a potential difference $V$ , it gains kinetic energy equal to the work done by the electric field: $K = qV$ , where $q$ is the charge of the particle. This kinetic energy is related to momentum by $K = \frac{p^2}{2m}$ , so $p = \sqrt{2mK} = \sqrt{2m q V}$ .

Therefore, the de Broglie wavelength becomes:
$\lambda = \frac{h}{\sqrt{2m q V}}$
For two different particles to have the same wavelength, the product $m q V$ must be the same for both (since $h$ is constant). This is the core idea — we don’t need to compute numerical wavelengths; we just compare the factors that determine them.

Identify the particles and their properties.

A proton has mass $m_p$ and charge $+e$ . An alpha particle (helium nucleus) has mass $4m_p$ and charge $+2e$ .

Write the condition for equal wavelengths.

For the proton accelerated through $V_1$ :

$\lambda = \frac{h}{\sqrt{2 m_p e V_1}}$
For the alpha particle accelerated through $V_2$ :
$\lambda = \frac{h}{\sqrt{2 (4m_p) (2e) V_2}}$
Setting them equal:
$\frac{h}{\sqrt{2 m_p e V_1}} = \frac{h}{\sqrt{2 \cdot 4m_p \cdot 2e \cdot V_2}}$

Cancel common factors. The $h$ and the factor $2$ cancel. We get:

$\frac{1}{\sqrt{m_p e V_1}} = \frac{1}{\sqrt{8 m_p e V_2}}$

Square both sides and solve for the ratio. Squaring: …
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TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
$Q. Which of the following gives proof of quantized electronic energy levels in hydrogen atom? (A) Atomic spectrum (B) Photoelectric effect (C) Emission of blackbody radiation (D) Davisson - Germer experiment$
›Reveal solutionSolution
Quantized energy levels in hydrogen mean electrons can occupy only specific, discrete energies. The atomic spectrum — sharp, discrete emission and absorption lines — directly proves this quantization, because each line corresponds to a transition between two fixed energy states. The answer is (A).

When an electron in hydrogen jumps from one energy level to another, it emits or absorbs a photon whose energy exactly matches the difference between those two levels. If energy levels were continuous (like a ramp), the atom could emit light of any wavelength, producing a continuous spectrum. Instead, hydrogen emits light at only certain precise wavelengths — the famous Balmer, Lyman, and Paschen series — each line a fingerprint of a specific quantum jump.

This is the smoking gun for quantization. The discrete lines in the atomic spectrum can only exist if the electron's allowed energies are themselves discrete.

Let's see why the other options don't demonstrate quantization in hydrogen specifically:

Atomic spectrum (A): The emission spectrum of hydrogen shows sharp lines at wavelengths given by the Rydberg formula,

$\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right),$
where $n_1$ and $n_2$ are integers. Each line corresponds to an electron transition between quantized levels $E_n = -\frac{13.6 \, \text{eV}}{n^2}$ . The photon energy is $E_{\gamma} = E_{n_2} - E_{n_1}$ , and because $n$ takes only integer values, the spectrum consists of discrete lines. This is direct, unambiguous evidence that the energy levels themselves are quantized.

Photoelectric effect (B): Einstein's explanation showed that light itself is quantized into photons of energy $E = h\nu$ , and that a minimum threshold frequency is needed to eject electrons from a metal surface. This proves the quantum nature of light, not the quantization of bound electronic energy levels in atoms. It tells us nothing about the internal structure of hydrogen. …
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TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$2.7 \times 10^{-2} \ \mathrm{kg \ mol^{-1}}$
›Reveal solutionSolution
Using the density formula for a crystal, we solve for the number of atoms per unit cell (Z). The calculation gives Z ≈ 4, so the correct option is (B).

Concept & Intuition

In solid-state physics, the density of a crystalline solid is directly related to how many atoms are packed into its unit cell. The formula
$\rho = \frac{Z \cdot M}{N_A \cdot a^3}$
connects density ( $\rho$ ), molar mass ( $M$ ), Avogadro’s number ( $N_A$ ), edge length ( $a$ ), and the number of atoms per unit cell ( $Z$ ). Here, we know everything except $Z$ , so we can rearrange and solve. The trick is to keep units consistent — especially converting picometers to meters.

Step-by-step solution

Write the density formula for a crystal For a cubic unit cell:

$\rho = \frac{Z \cdot M}{N_A \cdot a^3}$
where

$\rho$ = density (kg/m³),

$M$ = molar mass (kg/mol),

$N_A$ = Avogadro’s number (mol⁻¹),

$a$ = edge length (m),

$Z$ = number of atoms per unit cell (what we want).

Convert all quantities to SI units

Edge length: $a = 405 \ \text{pm} = 405 \times 10^{-12} \ \text{m} = 4.05 \times 10^{-10} \ \text{m}$

Molar mass: $M = 2.7 \times 10^{-2} \ \text{kg/mol}$

Density: $\rho = 2.7 \times 10^{3} \ \text{kg/m}^3$

Avogadro’s number: $N_A = 6.023 \times 10^{23} \ \text{mol}^{-1}$

Rearrange the formula to solve for $Z$

$Z = \frac{\rho \cdot N_A \cdot a^3}{M}$

Compute $a^3$

$a^3 = (4.05 \times 10^{-10})^3 = 4.05^3 \times 10^{-30} = 66.43 \times 10^{-30} \ \text{m}^3$
(Since $4.05^3 = 4.05 \times 4.05 \times 4.05 = 16.4025 \times 4.05 \approx 66.43$ )

Plug in the numbers

$Z = \frac{(2.7 \times 10^{3}) \times (6.023 \times 10^{23}) \times (66.43 \times 10^{-30})}{2.7 \times 10^{-2}}$

Simplify step by step

Cancel $2.7$ in numerator and denominator:

$Z = \frac{10^{3} \times 6.023 \times 10^{23} \times 66.43 \times 10^{-30}}{10^{-2}}$

Combine powers of 10: $10^{3} \times 10^{23} \times 10^{-30} = 10^{-4} …$
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TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$\dfrac{2h}{\pi}$
›Reveal solutionSolution

$L=\dfrac{2h}{\pi}=\dfrac{nh}{2\pi}$ gives $n=4$ , so $E=-\dfrac{13.6}{4^2}=-0.85$ eV — option (B).

Bohr's quantisation condition is $L=\dfrac{nh}{2\pi}$ . Setting it equal to the given angular momentum:

$\frac{2h}{\pi}=\frac{nh}{2\pi}\ \Rightarrow\ n=\frac{2h}{\pi}\times\frac{2\pi}{h}=4$ …
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TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$3~\text{GHz}$
›Reveal solutionSolution
The wavelength of an electromagnetic wave is found using $\lambda = c / f$ . With $f = 3 \times 10^9~\text{Hz}$ and $c = 3 \times 10^8~\text{m/s}$ , the wavelength is $0.1~\text{m}$ , so the correct option is (A).

The key idea is the universal wave relation for electromagnetic radiation in vacuum: speed = frequency × wavelength. Since all electromagnetic waves travel at the same speed $c$ in vacuum, knowing the frequency directly gives the wavelength.

Recall the fundamental wave equation

For any wave, $v = f \lambda$ , where $v$ is the wave speed, $f$ the frequency, and $\lambda$ the wavelength. For light in vacuum, $v = c = 3 \times 10^8~\text{m/s}$ .

Identify the given frequency

The problem states $f = 3~\text{GHz}$ . “Giga” means $10^9$ , so

$f = 3 \times 10^9~\text{Hz}.$

Solve for wavelength Rearranging $c = f \lambda$ gives

$\lambda = \frac{c}{f} = \frac{3 \times 10^8~\text{m/s}}{3 \times 10^9~\text{Hz}}.$

Simplify …
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TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$2.0 \times 10^{20} \ \text{s}^{-1}$
›Reveal solutionSolution

Power = (photons per second) $\times$ (energy per photon $=hc/\lambda$ ). Solving for $\lambda$ gives $3.978\times10^{-7}\,\text{m}=3978\,$ Å.

Solution

Each photon carries energy $E=\dfrac{hc}{\lambda}$ . If $N$ photons are emitted per second, the radiated power is
$P=N\cdot\frac{hc}{\lambda}\quad\Rightarrow\quad \lambda=\frac{N\,h\,c}{P}.$
Substitute $N=2.0\times10^{20}\,\text{s}^{-1}$ , $h=6.63\times10^{-34}\,\text{Js}$ , $c=3\times10^{8}\,\text{m s}^{-1}$ , $P=100\,\text{W}$ :

$\lambda=\frac{(2.0\times10^{20})(6.63\times10^{-34})(3\times10^{8})}{100}.$ …
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$_2$
›Reveal solutionSolution
The key is to count the total number of atoms, not just molecules. 6 g of H₂O contains 0.333 moles of molecules, but 1 mole of atoms (since each H₂O has 3 atoms). The only option that also contains exactly 1 mole of atoms is 0.4 g of He, which is 0.1 mole of He atoms — wait, that’s only 0.1 mole of atoms. Let’s check carefully: 6 g H₂O = 0.333 mol molecules × 3 atoms/molecule = 1.0 mol atoms. Among the options, 0.4 g He = 0.1 mol atoms (too few), 22 g CO₂ = 0.5 mol molecules × 3 = 1.5 mol atoms, 1 g H₂ = 0.5 mol molecules × 2 = 1.0 mol atoms, 12 g CO = 0.4286 mol molecules × 2 = 0.857 mol atoms. So the correct match is 1 g H₂. The answer is (C).

Concept & Intuition

This problem is a classic trap: students often compare masses or moles of molecules, but the question asks for the same number of atoms. Water (H₂O) has 3 atoms per molecule. So 6 g of water contains a certain number of molecules, but three times that many atoms. The trick is to convert everything to moles of atoms, not moles of molecules. We’ll compute the total moles of atoms in 6 g H₂O, then check each option for the same total.

Step-by-step solution

Find moles of atoms in 6 g H₂O

Molar mass of H₂O = 18 g/mol.

Moles of H₂O molecules = $\frac{6}{18} = \frac{1}{3}$ mol ≈ 0.333 mol.

Each H₂O molecule has 3 atoms (2 H + 1 O).

So total moles of atoms = $0.333 \times 3 = 1.0$ mol of atoms.

Check option (A): 0.4 g He

Helium is monatomic (1 atom per atom, obviously). Molar mass He = 4 g/mol.

Moles of He atoms = $\frac{0.4}{4} = 0.1$ mol.

That’s only 0.1 mol of atoms — far less than 1.0 mol.

✗ Not correct.

Check option (B): 22 g CO₂

Molar mass CO₂ = 44 g/mol.

Moles of CO₂ molecules = $\frac{22}{44} = 0.5$ mol.

Each CO₂ molecule has 3 atoms (1 C + 2 O).

Total moles of atoms = $0.5 \times 3 = 1.5$ mol.

✗ Not correct (too many).

Check option (C): 1 g H₂

Molar mass H₂ = 2 g/mol.

Moles of H₂ molecules = $\frac{1}{2} = 0.5$ mol.

Each H₂ molecule has 2 atoms.

Total moles of atoms = $0.5 \times 2 = 1.0$ mol.

✓ Exactly matches 1.0 mol of atoms.

Check option (D): 12 g CO …
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TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
$Q. If the velocity of a particle is reduced to half what is the percentage increase in its deBroglie wavelength? (A) 100 (B) 200 (C) 400 (D) 50$
›Reveal solutionSolution
de Broglie wavelength is inversely proportional to velocity; halving the velocity doubles the wavelength, which is a 100% increase.

The de Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$ , where $p = mv$ is the momentum. Since mass $m$ is constant, $\lambda \propto \frac{1}{v}$ . This inverse proportionality is the core idea: any change in velocity directly and oppositely affects the wavelength.

When the velocity is reduced to half, the wavelength becomes twice its original value. The question asks for the percentage increase, not the final value. A doubling corresponds to a 100% increase (from the original to twice the original, the increase is equal to the original itself).

Let’s work through it step by step.

Write the initial wavelength. Let the initial velocity be $v$ . The initial de Broglie wavelength is

$\lambda_1 = \frac{h}{mv}.$

Write the new wavelength after velocity is halved. New velocity $v_2 = \frac{v}{2}$ . Then

$\lambda_2 = \frac{h}{m(v/2)} = \frac{2h}{mv} = 2\lambda_1.$

Calculate the percentage increase. Percentage increase is defined as

$\frac{\lambda_2 - \lambda_1}{\lambda_1} \times 100\%.$
Substitute $\lambda_2 = 2\lambda_1$ :
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TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$2^{\text{nd}}$
›Reveal solutionSolution
The energy difference between the 2nd and 3rd Bohr orbits in hydrogen is $E_3 - E_2 = 1.89\ \text{eV}$ , so the correct choice is (C) 1.9 eV.

The key idea is that the energy levels of a hydrogen atom are quantized and given by the Bohr formula $E_n = -\frac{13.6\ \text{eV}}{n^2}$ . The minimum energy required to excite an electron from one orbit to a higher one is simply the difference between those two energy levels. No extra energy is needed for ionization or transitions to non‑integer levels — just the exact difference.

Recall the energy‑level formula for hydrogen In the Bohr model, the total energy of an electron in the $n$ -th orbit is

$E_n = -\frac{13.6\ \text{eV}}{n^2}$
The negative sign means the electron is bound; the ground state ( $n=1$ ) has the most negative energy.

Find the energy of the 2nd orbit

$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4\ \text{eV}$

Find the energy of the 3rd orbit

$E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.511\ \text{eV}$

Compute the energy difference The excitation energy from $n=2$ to $n=3$ is

$\Delta E = E_3 - E_2 = (-1.511) - (-3.4) = 1.889\ \text{eV}$
Rounded to one decimal place, this is $1.9\ \text{eV}$ . …
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Q.Which of the following will not show deflection from the path on passing through an electric field? Proton, cathode rays, electron, neutron.

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Q.Which of the following will not show deflection from the path on passing through an electric field?
Proton, cathode rays, electron, neutron.

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